Radial Distance of Projectile Shot from Earth's Surface

Click For Summary
SUMMARY

The discussion focuses on calculating the radial distance a projectile reaches when shot from Earth's surface at 0.758 times the escape speed. The relevant equations include the conservation of energy, expressed as K + U = 1/2 mv² - GmM/R = 0, where v = √(2GM/R). For part (a), the initial speed is adjusted to 0.758 times the escape speed, while part (b) requires recognizing the distinction between initial velocity and initial kinetic energy, leading to different calculations for maximum height.

PREREQUISITES
  • Understanding of gravitational potential energy and kinetic energy
  • Familiarity with the concept of escape velocity
  • Knowledge of the conservation of mechanical energy
  • Basic proficiency in algebra and physics equations
NEXT STEPS
  • Study the derivation of escape velocity from gravitational potential energy
  • Learn about the conservation of energy in gravitational fields
  • Explore the implications of varying initial speeds on projectile motion
  • Investigate the standard gravitational parameter (μ) and its applications
USEFUL FOR

Students preparing for physics exams, particularly those focusing on mechanics and gravitational theory, as well as educators seeking to clarify concepts related to projectile motion and energy conservation.

mariahkraft
Messages
6
Reaction score
0
A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE gives the radial distance (from the Earth's center) the projectile reaches if (a) its initial speed is 0.758 of the escape speed from Earth and (b) its initial kinetic energy is 0.758 of the kinetic energy required to escape Earth?

So basically what i got from lecture is that K+U=1/2mv2+-GmM/R=0
v=\sqrt{2GM/R}

for part a would I just multiply v=\sqrt{2GM/R} by .758 then plug that in for v in the 1/2 mv2?

then for part b would i multiply the entire 1/2mv2 by .758? Also one more thing for part b am i using my answer from part a for the speed or am i using standard escape speed?

I have my final on monday and I am trying to review for the test any help would be awesome Thanks!.
 
Physics news on Phys.org
mariahkraft said:
A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE gives the radial distance (from the Earth's center) the projectile reaches if (a) its initial speed is 0.758 of the escape speed from Earth and (b) its initial kinetic energy is 0.758 of the kinetic energy required to escape Earth?

So basically what i got from lecture is that K+U=1/2mv2+-GmM/R=0
v=\sqrt{2GM/R}

for part a would I just multiply v=\sqrt{2GM/R} by .758 then plug that in for v in the 1/2 mv2?

then for part b would i multiply the entire 1/2mv2 by .758? Also one more thing for part b am i using my answer from part a for the speed or am i using standard escape speed?

I have my final on monday and I am trying to review for the test any help would be awesome Thanks!.

Welcome to PF.

Your first equation is for escape velocity. But for escape velocity it is assumed that potential is 0 and kinetic energy is 0.

So what they are really giving you is 1/2m(.758v)2 + (-μ*m/Re) = PE at max height which is given by -μ*m/Rmax

The second part wants you to recognize the difference between initial velocity and initial KE which is v:v2

(Note I used μ as GM the standard gravitational parameter for Earth which numerically is 398K in SI units)
 
thank you!
 

Similar threads

Replies
40
Views
3K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K
  • · Replies 17 ·
Replies
17
Views
4K
Replies
1
Views
2K